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Lecture 11 Shuanglin Shao October 2nd and 7th, 2013 Matrix determinants: addition. Determinants: multiplication. Adjoint of a matrix. Cramer’s rule to solve a linear system. Recall that from the previous section, if a single row of A is multiplied by k, then for the resulting matrix, det(B) = k det(A). If the matrix B = kA, i.e., every entries in A are multiplied by the constant k, then we iterated the previous theorem det(B) = k n det(A). Example: det(A + B) 6= det(A) + det(B). Consider A= 1 2 2 5 ,B= 3 1 4 3 ?A + B == . 1 3 3 8 We have det(A) = 1, det(B) = 8, and det(A + B) = 23. Thus det(A + B) 6= det(A) + det(B). Theorem. Let A, B and C be n × n matrices that differ only in a single row, say the r -th row, and assume that the r -th row can be obtained by adding corresponding entries in the r -th row of A and B, then det(C ) = det(A) + det(B). This is proved by row expansion of determinants along the r -th row of C . Suppose that .. .. .. .. .. .. . . · · · . . . · · · . A= ar 1 ar 2 · · · arn ; B = br 1 br 2 · · · brn , .. .. .. .. .. .. . . ··· . . . ··· . and .. .. . . ··· a + b a + b ··· C = r 1 r 1 r 2 r 2 .. .. . . ··· where the rest entires are the same. .. . arn + brn , .. . We first observe that, if fixing the (r , j)-entry in either A, B or C , then the cofactors are the same, i.e., Crj is identical for A, B and C. det(C ) = = n X (arj + brj )Crj j=1 n X j=1 arj Crj + n X brj Crj j=1 = det(A) + det(B). Note that A + B is not the same as C : A + B is obtained by adding the corresponding entries. Example. 1 7 5 2 0 3 1 + 0 4 + 1 7 + (−1) 1 7 5 = 2 0 3 1 4 7 1 7 5 + 2 0 3 0 1 −1 . Determinants of a product. Lemma 2.3.2. If B is an n × n matrix and E is an n × n elementary matrix, then det(EB) = det(E ) det(B). Let E be an elementary matrix. If we multiply B by E from the left, EB is the matrix obtained by performing an elementary row operation on B. Case 1. If E is obtained from In by multiplying k to a row, then EB is the matrix that is obtained by multiplying k to the same row. det(EB) = k det(B) = det(E ) det(B) because det(E ) = k. Case 2. If E is obtained from In by exchanging two rows, then EB is the matrix that is obtained by exchanging the same two rows: det(EB) = − det(B) = det(E ) det(B) because det(E ) = −1. Case 3. If E is obtained from In by adding k times a row to another row, then EB is the matrix that is obtained by adding k times the same row to the same another row: det(EB) = det(B) = det(E ) det(B) because det(E ) = 1. Remark. If B is an n × n matrix and E1 , E2 , · · · , Er are n × n elementary matrices, then det(E1 E2 · · · Er B) = det(E1 ) det(E2 ) · · · det(Er ) det(B). Determinant test for invertibility Theorem. A square matrix A is invertible if and only if det(A) 6= 0. Proof. If A is invertible, then A is expressed as a product of elementary matrices: A = E1 E2 · · · Er . Then det(A) = det(E1 ) det(E2 ) · · · det(Er ). Since for each elementary matrix E , det(E ) 6= 0. Thus det(A) 6= 0. Cont. On the other hand, suppose det(A) 6= 0. We apply elementary row operations to reduce A to reduced row echelon form R. That is to say, R = E1 E2 · · · Er A. Then det(R) = det(E1 ) det(E2 ) · · · det(Er ) det(A) 6= 0. This implies det(R) 6= 0. Since R is a square matrix and is in reduced row echelon form, R is the identity matrix. In other words, A reduces to In after a series of elementary row operations. Hence A is invertible. 1 2 3 Example. Let A = 1 0 1 . Since the first and third rows of 2 4 6 A is proportional, det(A) = 0. Hence A is not invertible. If A and B are square matrices of the same size, then det(AB) = det(A) det(B). Case 1. If A is not invertible, det(A) = 0. It also follows that AB is not invertible, which implies that det(AB) = 0. Thus det(AB) = 0 = det(A) B . Case 2. If A is invertible, then A is a product of elementary matrices, E1 , E2 , · · · , Er : A = E1 E2 · · · Er B. Thus det(AB) = det(E1 E2 · · · Er B) = det(E1 ) det(E2 ) · · · det(Er ) det(B). Since det(A) = det(E1 ) det(E2 ) · · · det(Er ), det(AB) = det(A) det(B). Consider A = 3 1 2 1 ,B= −1 3 5 8 2 17 3 14 . The product AB = . We verify that det(A) = 1, det(B) = −23, det(AB) = −23. Thus det(AB) = det(A) det(B). A corollary to the theorem is Theorem. If A is invertible, then det(A−1 ) = 1 . det(A) This is because In = A × A−1 . Thus 1 = det(In ) = det(A) det(A−1 ). Def. If A is any n × n matrix, and Cij is the cofactor of aij , then the matrix C11 C21 · · · Cn1 C12 C22 · · · Cn2 .. .. .. . . ··· . C1n C2n · · · Cnn is called the adjoint of the matrix A, denoted by adj(A) . Recall that Cij = (−1)i+j Mij , and Mij is the determinant of the (n − 1) × (n − 1) matrix obtained from A by removing the i-th row and the j-th column of A. Example. Let 3 2 −1 3 . A= 1 6 2 −4 0 The adjoint of A is 12 4 12 2 −10 . adj(A) = 6 −16 16 16 We compute 2 −1 4 0 = 4. 2+1 C21 = (−1) Inverse of a matrix using its adjoint. Theorem. If A is an invertible matrix, then A−1 = 1 adj(A). det(A) Proof. We show that A adj(A) = det(A)In . Let B = A adj(A) = [bij ]. The (i, j)-entry bij of the product matrix A adj(A) is coming from the i-th row of A and the j-th column of adj(A). Note that the j-th column of adj(A) is the cofactor of the entry ajk of the matrix A, 1 ≤ k ≤ n. bij = n X k=1 aik Cjk = ai1 Cj1 + ai2 Cj2 + · · · + ain Cjn . Fixing i. We discuss two cases. Case 1. When j = i, by the definition of determinants of matrices, bij = det(A). Case 2. When j 6= i, we prove that bij = 0. Suppose that i < j; the proof for i > j is similar. We construct a matrix that differs from the matrix A only in the j-th row: the j-th row is the same as the i-th row. .. .. .. . ··· . . ai1 ai2 · · · ain .. .. . D = ... . ··· . ai1 ai2 · · · ain .. .. .. . . ··· . Since D contains two identical rows, det(D) = 0. We expand the determinant along the j-th row: 0 0 det(D) = 0 = ai1 Cj1 + ai2 Cj2 + · · · ain Cjn0 . Since D only differs from A in the j-th row, Cjk0 = Cjk . So bij = 0, for i 6= j. Hence the matrix B is det(A)In . This proves the theorem. Using the adjoint to find an inverse matrix. 3 2 −1 3 . From the previous example, Example. A = 1 6 2 −4 0 the adjoint of A is computed. By the theorem, the inverse of A is 12 4 12 1 6 2 −10 . det(A) −16 16 16 The determinant of A, det(A) = 64. So the inverse is 12 4 12 64 6 64 − 16 64 64 2 64 16 64 64 . − 10 64 16 64 Cramer’s rule. Theorem. If Ax = b is a system of n linear equations in n unknowns such that det(A) 6= 0, then the system has a unique solution. This solution is x1 = det(A2 ) det(An ) det(A1 ) , x2 = , · · · , xn = det(A) det(A) det(A) when Aj is the matrix obtained by replacing the entries in the j-th column of A by the entires in the matrix b1 b2 b = . . . . bn If det(A) 6= 0, then A is invertible. We obtain C11 C21 C 1 12 C22 x = A−1 b = . .. det(A) .. . C1n C2n multiply Ax = b by A−1 to ··· ··· ··· ··· Cn1 Cn2 .. . Cnn b1 b2 .. . . bn We multiply the matrices on the right hand side out to obtain the i-th entry. b1 C1i + b2 C2i + · · · + bn Cni . This is precisely the determinant of the following matrix expanding along the i-th column: a11 · · · b1 · · · a1n a21 · · · b2 · · · a2n .. .. .. .... . . . .. an1 · · · bn · · · ann where the i-th column of A is replaced by the column vector. So xi = det(Ai ) . det(A) Using Cramer’s rule to solve a linear system. Use the Cramer’s rule to solve x1 + 2x3 −3x1 + 4x2 + 6x3 −x1 − 2x2 + 3x3 = 6, = 30, = 8. 1 0 2 6 0 2 A = −3 4 6 , A1 = 30 4 6 −1 −2 3 8 −2 3 and 1 6 2 1 0 6 A2 = −3 30 6 , A3 = −3 4 30 −1 8 3 −1 −2 8 The complexity here is to compute the determinants here. Therefore x1 = det(A1 ) −10 det(A2 ) 18 = , x2 = = . det(A) 11 det(A) 11 and x3 = det(A3 ) 38 = . det(A) 11 We usually use the elementary row operations to compute the determinants of matrices. Homework and Reading. Homework. Ex. #4, #6,#8, #10, #16, #18, #20,#24, #30, and the True-False exercise on page 116. Reading. Section 3.1.